Is ${707875}$ divisible by $3$ ?
Explanation: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {707875}= &&{7}\cdot100000+ \\&&{0}\cdot10000+ \\&&{7}\cdot1000+ \\&&{8}\cdot100+ \\&&{7}\cdot10+ \\&&{5}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {707875}= &&{7}(99999+1)+ \\&&{0}(9999+1)+ \\&&{7}(999+1)+ \\&&{8}(99+1)+ \\&&{7}(9+1)+ \\&&{5} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {707875}= &&\gray{7\cdot99999}+ \\&&\gray{0\cdot9999}+ \\&&\gray{7\cdot999}+ \\&&\gray{8\cdot99}+ \\&&\gray{7\cdot9}+ \\&& {7}+{0}+{7}+{8}+{7}+{5} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${707875}$ is divisible by $3$ if ${ 7}+{0}+{7}+{8}+{7}+{5}$ is divisible by $3$ Add the digits of ${707875}$ $ {7}+{0}+{7}+{8}+{7}+{5} = {34} $ If ${34}$ is divisible by $3$ , then ${707875}$ must also be divisible by $3$ Add the digits of ${34}$ $ {3}+{4} = \color{#9D38BD}{7} $ If $\color{#9D38BD}{7}$ is divisible by $3$ , then ${34}$ must also be divisible by $3$ $\color{#9D38BD}{7}$ is not divisible by $3$, therefore ${707875}$ must not be divisible by $3$.